Orthogonality of modes and Rayleigh
quotient
• Orthogonality of continuum modes and the
relationship to the discrete case
• The Rayleigh quotient
– What does it have to do with orthogonality?
• Example of application to beam with end
mass
– How do we select approximations to modes?
8.5: Orthogonality of modes
d ⎡ dYr ⎤
2
− ⎢T
=
ω
r ρYr
⎥
dx ⎣ dx ⎦
• For string vibration
• Multiply equation by Ys and integrate
−∫
L
0
d ⎡ dYr ⎤
dYr
T
Ys dx = − Y sT
⎢
⎥
dx ⎣ dx ⎦
dx
L
0
dYr dYs
+∫ T
dx
0
dx dx
L
• So (how did we get rid of integrated term?)
L
L
dYr dYs
2
2
r
s dx = ωs ∫ ρ YY
r
s dx
∫0 T dx dx dx = ωr ∫0 ρ YY
0
L
• For distinct vibration frequencies
∫
L
0
ρ YY
r
s dx = δ rs
dYr dYs
2
T
dx
=
ω
r δ rs
∫0 dx dx
L
Connection to discrete case
• For continuum problem
∫
L
0
ρ YY
r
s dx = δ rs
dYr dYs
2
T
dx
=
ω
r δ rs
∫0 dx dx
L
• For discrete problem
yTr My s = δ rs
yTr Ky s = ωr2δ rs
• What is the
connection?
• For beam vibrations
∫
L
0
mYY
r
s dx = δ rs
∫
L
0
d 2Yr d 2Ys
2
=
ω
EI 2
dx
r δ rs
2
dx dx
Expansion theorem
• With orthogonality we have following expansion
theorem for beams: Any function Y(x)
representing a possible displacement of the
beam, which implies that Y(x) satisfies the
boundary conditions for the beam and is C2
continuous can be expanded in the absolutely
and uniformly convergent series of
eigenfunctions (Do you know what these terms
mean?)
∞
Y (x) = ∑ crYr (x)
r =1
L
cr = ∫ m(x)Yr (x)Y (x)dx
0
8.8: Rayleigh quotient: The variational
approach
• We had
∫
L
0
• So we can write
• In general, with an
approximation to a vibration
mode
• As in the algebraic case, the
error in the frequency is like
the square of error in mode
• In general
Vmax
2
ω =
Tmax
∫
ρ YY
r
s dx = δ rs
L
0
T
dYr dYs
dx = ωr2δ rs
dx dx
2
dY
ω = ∫ T ⎡⎢ r ⎤⎥ dx
0
⎣ dx ⎦
2
r
L
2
⎡ dY ⎤ dx
T
∫0 ⎢⎣ dx ⎥⎦
2
ω =
L
2
Y
dx
ρ
∫
L
0
Problem 8.29
• Beam with end
mass
• Estimate natural
frequency using
Y(x)=2-3(x/L)+(x/L)3
for M=0.5mL
• For this problem
2
⎡ ∂2 y ⎤
1 L
V = ∫ EI ⎢ 2 ⎥ dx
2 0
⎣ ∂x ⎦
2
1 L ⎡ ∂y ⎤
1 ⎡ ∂y(0, t) ⎤
T = ∫ m ⎢ ⎥ dx + M ⎢
2 0 ⎣ ∂t ⎦
2 ⎣ ∂t ⎥⎦
y(x, t) = Y (x)cos(ω t − φ )
How did we select Y?
2
Calculating the Rayleigh quotient
• Potential energy
V (t ) =
Vmax
• Kinetic energy
2
⎡∂ y ⎤
1
2
=
EI
dx
V
cos
(ω t − φ )
max
⎢
2 ⎥
∫
0
2
⎣ ∂x ⎦
L
2
2
2
⎡∂ Y ⎤
1
EI L ⎛ 6x ⎞
6EI
= ∫ EI ⎢ 2 ⎥ dx =
=
dx
⎜ 3⎟
∫
0
∂
2 0
x
2
L ⎠
L3
⎝
⎣
⎦
L
2
2
2
1 L ⎡ ∂y ⎤
1 ⎡ ∂y(0, t) ⎤
=
T = ∫ m ⎢ ⎥ dx + M ⎢
⎥
0
2
2 ⎣ ∂t ⎦
⎣ ∂t ⎦
ω2 ⎡ L
2
2
⎤ sin2 (ω t − φ ) = ω 2T sin2 (ω t − φ )
(0)
+
m
Y
dx
MY
ref
⎢∫
⎥
2 ⎣
⎦
0
L
Tref = ∫ m Y 2dx + MY 2 (0)
Checks?
0
3 2
⎧
⎫ 103
1⎪ L⎡
x ⎛x⎞ ⎤
2⎪
= ⎨m ∫ ⎢2 − 3 + ⎜ ⎟ ⎥ dx + 0.5mL(2) ⎬ =
mL
0
2⎪
L ⎝L⎠ ⎦
⎣
⎪⎭ 70
⎩
•Altogether
ω = 2.0193
EI
mL4
ωexact = 2.0164
EI
mL4
Reading assignment
Sections 8.9, 8.10
Source: www.library.veryhelpful.co.uk/ Page11.htm